# FINAL EXAM SOLUTIONS Math 21a, Spring 03

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2 or every c, the function u(x, t) = (2 cos(ct) + 3 sin(ct)) sin(x) is a solution to the wave equation u tt = c 2 u xx. rue Just differentiate. he length of the curve r(t) = (t, sin(t)), where t [, 2π] is 1 + cos 2 (t) dt. 2π rue he speed at time t is r (t) = 1 + cos 2 (t). Let (x, y ) be the maximum of f(x, y) under the constraint g(x, y) = 1. hen f xx (x, y ) <. alse. While this would be true for g(x, y) = f(y), where the constraint is a straight line parallel to the y axes, it is false in general. he function f(x, y, z) = x 2 y 2 z 2 decreases in the direction (2, 2, 2)/ 8 at the point (1, 1, 1). alse. It increases in that direction. Assume is a vector field satisfying (x, y, z) 1 everywhere. or every curve C : r(t) with t [, 1], the line integral C dr is less or equal than the arc length of C. rue. r r r. Let be a vector field which coincides with the unit normal vector N for each point on a curve C. hen C dr =. rue. he vector field is orthogonal to the tangent vector to the curve. he divergence of the gradient of any f(x, y, z) is always zero. alse. div(grad(f)) = f is the Laplacian of f. or every function f, one has div(curl(grad(f))) =. rue. Both because div(curl( ) = and curl(grad(f) =. If for two vector fields and G one has curl( ) = curl( G), then = G + (a, b, c), where a, b, c are constants. alse. One can also have = G + grad(f) which are vectorfields with the same curl. or every vector field the identity grad(div( )) = holds. alse. = (x 2, y 2, z 2 ) has div( ) = (2x, 2y, 2z) which has a nonzero gradient. If a nonempty quadric surface g(x, y, z) = ax 2 + by 2 + cz 2 = 5 can be contained inside a finite box, then a, b, c. rue. If one or two of the constants a, b, c are negative, we have a hyperboloid which all can not be contained into a finite space. If all three are negative, then the surface is empty. 2

3 If is a vector field in space then the flux of through any closed surface S is. alse. While it is true that the flux of curl( ) vanishes through every closed surface, this is not true for itself. ake for example = (x, y, z). If div( )(x, y, z) = for all (x, y, z), then curl( ) = (,, ) for all (x, y, z). alse. ake ( y, x, ) for example. he flux of the vector field (x, y, z) = (y + z, y, z) through the boundary of a solid region E is equal to the volume of E. alse. By the divergence theorem, the flux through the boundary is E div( ) dv but div( ) =. So the flux is zero. If in spherical coordinates the equation φ = α (with a constant α) defines a plane, then α = π/2. rue. Otherwise, it is would be a cone (or for α = or α = π a half line). or every function f(x, y, z), there exists a vector field such that div( ) = f. rue. In order to solve P x + Q y + R z = f just take = (,, z f(x, y, w) dw). 3

4 Problem 2) (1 points) Match the equations with the objects. No justifications are needed. I II III IV V VI VII VIII Enter I,II,III,IV,V,VI,VII,VIII here Equation V g(x, y, z) = cos(x) + sin(y) = 1 VIII y = cos(x) sin(x) I r(t) = (cos(t), sin(t)) II r(u, v) = (cos(u), sin(v), cos(u) sin(v)) VI (x, y, z) = (cos(x), sin(x), 1) IV z = f(x, y) = cos(x) + sin(y) VII g(x, y) = cos(x) sin(y) = 1 III (x, y) = (cos(x), sin(x)) Problem 3) (1 points) Mark with a cross in the column below conservative if a vector fields is conservative (that is if curl( )(x, y, z) = (,, ) for all points (x, y, z)). Similarly, mark the fields which are incompressible (that is if div( )(x, y, z) = for all (x, y, z)). No justifications are needed. 4

5 Vectorfield conservative incompressible curl( ) = div( ) = (x, y, z) = ( 5, 5, 3) X X (x, y, z) = (x, y, z) X (x, y, z) = ( y, x, z) (x, y, z) = (x 2 + y 2, xyz, x y + z) (x, y, z) = (x 2yz, y 2zx, z 2xy) X Problem 4) (1 points) Let E be a parallelogram in three dimensional space defined by two vectors u and v. a) (3 points) Express the diagonals of the parallelogram as vectors in terms of u and v. Solution: first diagonal u + v, second diagonal u v. b) (3 points) What is the relation between the length of the crossproduct of the diagonals and the area of the parallelogram? Solution: (u + v) (u v) = 2v u = 2 times area of of parallelogram. c) (4 points) Assume that the diagonals are perpendicular. What is the relation between the lengths of the sides of the parallelogram? Solution: (u + v) (u v) = u 2 v 2 =, so that u = v. Problem 5) (1 points) ind the volume of the largest rectangular box with sides parallel to the coordinate planes that can be inscribed in the ellipsoid x2 + y2 + z2 = Solution. he volume of the box is 8xyz. he Lagrange equations are 8yz = λx/2 8xz = λ2y/9 8xy = λ2z/25 x y2 9 + z = We can solve this by solving the first three equations for λ and expressing y, z by x, plugging this into the fourth equation. An other way to solve this is to multiply the first equation with x, the second with y and third with z. 5

6 he solution is x = 2/ 3, y = 3, z = 5/ 3. he maximal volume is 8xyz = 8/ 3. Problem 6) (1 points) Evaluate 8 2 y 1/3 y 2 e x2 x 8 dxdy. Solution: his type II integral can not be computed as it is. We write it as a type I integral: from the boundary relation x = y 1/3 we obtain y = x 3 and y = 8 corresponds to x = 2: he result is e x 3 2 x x e x2 x 8 3 ex2 y 2 e x2 x 8 dx dx dydx e x2 /6 2 = (e 4 1)/ Problem 7) (1 points) (x y) 4 (x+y) 4 In this problem we evaluate D x and y axes and the line x + y = 1. dxdy, where D is the triangular region bounded by the a) (3 points) ind the region R in the uv-plane which is transformed into D by the change of variables u = x y, v = x + y. (It is enough to draw a carefully labeled picture of R.) b) (3 points) ind the Jacobian (x,y) (u,v) c) (4 points) Evaluate D (x y) 4 (x+y) 4 Solution. a) ake R with the edges (, ), (1, 1) and ( 1, 1). b) he Jacobian is x u y v x v y u = (1/4 + 1/4) = 1/2. c) It is best evaluated as a type II integral: 1 1 v 2 v u4 /v 4 dudv = v/5 = dv = 1/1. of the transformation (x, y) = ( u+v 2, v u 2 ). dxdy using the above defined change of variables. Problem 8) (1 points) 6

7 a) (3 points) ind all the critical points of the function f(x, y) = (x 4 8x 2 + y 2 + 1). b) (3 points) Classify the critical points. c) (2 points) Locate the local and absolute maxima of f. d) (2 points) ind the equation for the tangent plane to the graph of f at each absolute maximum. Solution. a) (±2, ) and (, ). b) ( 2, ) is a local maximum with value 15. (, ) is a saddle with value 1. (2, ) is a maximum with value 15. c) he local maxima are (±2, ). hey are also the absolute maxima because f decays at infinity. d) o calculate the tangent plane at the maximum, write the graph of f as a level surface g(x, y, z) = z f(x, y). he gradient of g is orthogonal to the surface. We have g = (,, 1) so that the tangent plane has the equation z = d = const. Plugging in the point (±2,, 15) shows that z = 15 is the equation for the tangent plane for both maxima. Problem 9) (1 points) ind the volume of the wedge shaped solid that lies above the xy-plane and below the plane z = x and within the cylinder x 2 + y 2 = 4. Solution. Use polar coordinates and note that the wedge is above the right side of the unit disc: 2 π/2 π/2 he solution is 16/3. r 2 cos(θ) dθdr = 16/3 Problem 1) (1 points) Let the curve C be parametrized by r(t) = (t, sin t, t 2 cos t) for t π. Let f(x, y, z) = z 2 e x+2y + x 2 and = f. ind C d r. 7

8 Solution. Use the fundamental theorem of line integrals. he result is f(r(π)) f(r()) = f(π,, π 2 ) f(,, ) = π 4 e π + π 2 = π 4 e π + π 2. Problem 11) (1 points) Evaluate the line integral of the vector field (x, y) = (y 2, x 2 ) in the clockwise direction around the triangle in the xy-plane defined by the points (, ), (1, ) and (1, 1) in two ways: a) (5 points) by evaluating the three line integrals. b) (5 points) using Greens theorem. Solution. he problem asks to do this in the clockwise direction. We do it in the counterclockwise direction and change then the sign. a) 1 (t, ) (1, ) dt + 1 (1, t) (, 1) dt + 1 (1 t, 1 t) ( 1, 1) dt = + 1 2/3 = 1/3. So, the result for the clockwise direction is 1/3. b) he curl of is 2x 2y. 1 x (2x 2y)dy dx = So, the result for the clockwise direction is 1/3. 1 2x 2 x 2 dx = 1/3 Problem 12) (1 points) Use Stokes theorem to evaluate the line integral of (x, y, z) = ( y 3, x 3, z 3 ) along the curve r(t) = (cos(t), sin(t), 1 cos(t) sin(t)) with t [, 2π]. Solution. he curve is contained in the graph of the function f(x, y) = 1 x y which is parameterized by r(u, v) = (u, v, 1 u v) and has the normal vector r u r v = (1,, 1) (, 1, 1) = (1, 1, 1). he curl of is (,, 3x 2 + 3y 2 ) so that (r(u, v)) (r u r v ) = 3(x 2 + y 2 ). he surface is parameterized over the region R = {u 2 + v 2 1} and S ds = 2π 3r 3 dθdr = 3π. 2 1 Problem 13) (1 points) Let S be the graph of the function f(x, y) = 2 x 2 y 2 which lies above the disk {(x, y) x 2 + y 2 1} in the xy-plane. he surface S is oriented so that the normal vector points upwards. 8

9 Compute the flux S ds of the vectorfield through S using the divergence theorem. = ( 4x + x2 + y y 2, 3y, 7 z 2xz 1 + 3y 2 ) Solution. We apply the divergence theorem to the region E = { z f(x, y), x 2 + y 2 1 }. Using div( ) = 2, we get div( ) dv = 1 2π 2 r 2 = ( 2) 1 2π ( 2) r drdθdz (2 r 2 ) r drdθdz = ( 2)(2π)(2/2 1/4) = 3π. By the divergence theorem, this is the flux of through the boundary of E which consists of the surface S, the cylinder S 1 : r(u, v) = (cos(u), sin(u), v) with normal vector r u r v = ( sin(u), cos(u), ) (,, 1) = (cos(u), sin(u), ) plus the flux through the floor S 2 : r(u, v) = (v sin(u), v cos(u), ) with normal vector r u r v = (,, v). he flux through S 1 is he flux through S 2 is S1 ds = 1 = 2π 1 2π S2 ds = 1 = 1 2π (cos(u), sin(u), v) (cos(u), sin(u), ) dudv ( 4 cos 2 (u) + 3 sin 2 (u)) dudv = π 2π By the divergence theorem, S ds + S 1 3π + π + 7π = 5π. (v sin(u), v cos(u), 7) (,, v) dudv ( 7v) dudv = 7π ds + S 2 ds = 3π so that S ds = 9

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